Finding the perimeter of a right triangle with area given I have a question where the teacher wants me to find the perimeter of a right triangle using the area which is equal to $1cm^2$.  I've tried a few approaches but seem to be struggling and I can't even find the side lengths.  is this even possible?
would love some help thanks
 A: If that is all the information you have, you can't find the perimeter.  If the triangle is isosceles and the leg is $x$, the area is $\frac {x^2}2$, so $x=\sqrt 2$ and the perimeter is $2\sqrt 2+2$.  If the triangle has a very long leg $L$ and a short leg of length $x$, the area is $\frac 12xL$, so $L=\frac 2x$.  The perimeter is then $x+L+\sqrt{x^2+L^2}\approx 2L$ which can be as large as you like.
A: Let $ l, b $ be the side lengths. Then,
$$\dfrac {1}{2} lb=1 $$
So,
$$l=\dfrac {2}{b} $$
The perimeter is
$$ l+b+\sqrt {l^2+b^2} = \dfrac {2}{b}+b+\sqrt {\dfrac {4}{b^2}+b^2}$$
You should be able to simplify this expression. 
A: If the base is $b$ and the height is $c$ then the area is $bc/2$ and the perimeter is $b+c+\sqrt{b^2+c^2}$.
If $bc/2 = 1$ (sq. cm.) then $c = 2/b$, so the perimeter is $\displaystyle b+\frac 2 b + \sqrt{b^2 + \left(\frac 2 b\right)^2}$.
As $b$ goes from $0$ to $\sqrt{2}$, the perimeter goes from $\infty$ down to $2+2\sqrt{2}$, and as $b$ goes from $\sqrt{b}$ to $\infty$, the perimeter goes from $2+2\sqrt{2}$ back up to $\infty$.
So the smallest that the perimeter could be is $2+2\sqrt{2}$, but it could be anything bigger.
